Dirac delta function proof

Question

I got the function

$$f(x)=e^{-|x|}$$

I want to show that $f''(x)=f(x)-2\delta(x)$ where $\delta(x)$ is the Dirac delta function.

I know that I can solve it with a known theorem but can I prove it without using it?

Answer 1

We have that $f$ satisfies the following integral equation: $$f'(x)-\int_{0}^x f=-\operatorname{sgn}(x)=-2u(t)+1$$ Which is equivalent to the given differential equation.

Answer 2

Here is a heuristic answer:

\begin{align*} f''(x) &= \bigl( e^{-|x|} \bigr)'' \\ &= \bigl( - \operatorname{sgn}(x)e^{-|x|} \bigr)' \\ &= -2\delta(x)e^{-|x|} + (\operatorname{sgn}(x))^2 e^{-|x|} \\ &= -2\delta(x) + f(x). \end{align*}

For a more rigorous proof, let $\varphi \in \mathcal{D}(\mathbb{R})$ be any compactly supported smooth function. Since we are dealing with derivatives of distributions, we may invoke the weak derivative to reformulate the question. Indeed, it is equivalent to showing that

$$ \int_{\mathbb{R}} f(x)(\varphi(x) - \varphi''(x)) \, \mathrm{d}x = 2\varphi(0). $$

Now, by performing integration by parts twice,

\begin{align*} \int_{\mathbb{R}} f(x) \varphi''(x) \, \mathrm{d}x &= \int_{-\infty}^{0} e^{x} \varphi''(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi''(x) \, \mathrm{d}x \\ &= \underbrace{\left[ e^{x} \varphi'(x) \right]_{-\infty}^{0} + \left[ e^{-x} \varphi'(x) \right]_{0}^{\infty}}_{=0} - \int_{-\infty}^{0} e^{x} \varphi'(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi'(x) \, \mathrm{d}x \\ &= \left[ -e^{x} \varphi(x) \right]_{-\infty}^{0} + \left[ e^{-x} \varphi(x) \right]_{0}^{\infty} + \int_{-\infty}^{0} e^{x} \varphi(x) \, \mathrm{d}x + \int_{0}^{\infty} e^{-x} \varphi(x) \, \mathrm{d}x \\ &= -2\varphi(0) + \int_{\mathbb{R}} f(x) \varphi(x) \, \mathrm{d}x. \end{align*}

Therefore the desired claim follows.

Answer 3

Let $f(x)=e^{-|x|}.$ $$\frac{d |x|}{dx}=\text{sgn}(x)=2 \theta(x)-1,$$ Where $$\theta(x)=1, x>0;~ 0, x\le 0.$$ Further, $$\frac{d \theta(x)}{dx}=\delta(x).$$ So $$f'(x)=\frac{d e^{-|x|}}{dx}=-e^{-|x|} \text{sig}(x)=-e^{-|x|}~[2\theta(x)-1]$$ $$f''(x)=e^{-|x|}~\text{sgn}^2(x)-e^{-|x|}\frac{d~\text{sgn}(x)}{dx}=f(x)-2f(x) \frac{\theta(x)}{dx}=f(x)-2\delta(x) f(x).$$ Note that $\text{sgn}^2(x)=1.$ Next, we can write $$f''(x)=f(x)-2\delta(x) f(0)=f(x)-2\delta(x)$$